How do you sketch the graph #y=x^4-2x^3+2x# using the first and second derivatives?

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Answer 1 · 2018-06-12T10:37:51

See process below

1.- Domain is #RR#
2.- Analyze the roots of equation #x^4-2x^3+2x=0# in order to find axis intercepts

Notice that #x^4-2x^3+2x=x(x^3-2x^2+2)=0# thus #(0,0)# is a passing point. There is no more integer roots

3.- First derivative #f´(x)=4x^3-6x^2+2=(x-1)^2(4x+2)#

Analyze the sign of derivative. #(x-1)^2# is allways positive, then the sign of derivative dependes only of sign of #4x+2#

#4x+2>=0#
#x>=-1/2#

Then function is increasing in #(-1/2,oo)# and decreasing in #(-oo,-1/2)# then in #x=-1/2# has a minimum

Derivative is zero in #x=-1/2# and #x=1# but #x=1# is not a maximum neither a minimum because

#f´´(x)=12x^2-12x=12x(x-1)#. then #f´´# become #0# only in #x=0# and #x=1#. Thus inflection points.

A sketch is presented below
enter image source here